Find the position of centre of gravity of an unequal angle section 10 cm* 16 cm * 2 cm
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A Textbook of Engineering Mechanics Let x and y be the co-ordinates of the centre of gravity with respect to some axis of reference, then 1 1 2 2 3 3 1 2 3 ........a x a x a x x a a a + + + = + + and 1 1 2 2 3 3 1 2 3 ........ ... a y a y a y y a a a + + + = + + + where a1, a2, a3........ etc., are the areas into which the whole figure is divided x1, x2, x3 ..... etc., are the respective co-ordinates of the areas a1, a2, a3....... on X-X axis with respect to same axis of reference. y1, y2, y3....... etc., are the respective co-ordinates of the areas a1, a2, a3....... on Y-Y axis with respect to same axis of the reference. Note. While using the above formula, x1, x2, x3 ..... or y1, y2, y3 or x and y must be measured from the same axis of reference (or point of reference) and on the same side of it. However, if the figure is on both sides of the axis of reference, then the distances in one direction are taken as positive and those in the opposite directions must be taken as negative. 6.8. CENTRE OF GRAVITY OF SYMMETRICAL SECTIONS Sometimes, the given section, whose centre of gravity is required to be found out, is symmetrical about X-X axis or Y-Y axis. In such cases, the procedure for calculating the centre of gravity of the body is very much simplified; as we have only to calculate either x or y . This is due to the reason that the centre of gravity of the body will lie on the axis of symmetry. Example 6.1. Find the centre of gravity of a 100 mm × 150 mm × 30 mm T-section. Solution. As the section is symmetrical about Y-Y axis, bisecting the web, therefore its centre of gravity will lie on this axis. Split up the section into two rectangles ABCH and DEFG as shown in Fig 6.10. Let bottom of the web FE be the axis of reference. (i) Rectangle ABCH a1 = 100 × 30 = 3000 mm2
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